Staff: Mentor

Why? It's just a consequence of the definition of the Compton wavelength, which was defined as a convenience for analyzing experiments where photons scatter off electrons (Compton was one of the first to do them).

No, it doesn't. First, your value of ##r## doesn't make sense, because ##r## is the distance of the particle from some arbitrarily chosen point, not a wavelength or half wavelength. Second, your result is ##h / 2##, but the spin of an electron is ##\hbar / 2##, where ##\hbar = h / 2 \pi##. And the spin of a photon is just ##\hbar##. Your result doesn't match either of those.

The formulas imply it calculates the spin of the electron "as if" the electron is a photon with the said wavelength.

It implies no such thing. If anything, it implies that you are thinking of the electron as a little spinning sphere with radius equal to half of its Compton wavelength, and calculating the orbital angular momentum of a point on the sphere's surface. And getting an answer that is not the same as the electron's intrinsic angular momentum.